$g(t) = 2t^{2}+4(f(t))$ $h(n) = 3n-1+2(f(n))$ $f(t) = -3$ $ g(f(-1)) = {?} $
Answer: First, let's solve for the value of the inner function, $f(-1)$ . Then we'll know what to plug into the outer function. $f(-1) = -3$ $f(-1) = -3$ Now we know that $f(-1) = -3$ . Let's solve for $g(f(-1))$ , which is $g(-3)$ $g(-3) = 2(-3)^{2}+4(f(-3))$ To solve for the value of $g$ , we need to solve for the value of $f(-3)$ $f(-3) = -3$ $f(-3) = -3$ That means $g(-3) = 2(-3)^{2}+(4)(-3)$ $g(-3) = 6$